# Prove that, $a_n \cdot b_n \to a \cdot b$

I want to prove this exercise:

Let $a_n \to a$ and $b_n \to b$ for $n \to \infty$ Prove that, $a_n \cdot b_n \to a \cdot b$.

My proof:

Let $\epsilon > 0$ and $N_1 \in \mathbb{N}$ such that $|a_n - a| < \epsilon \forall n \geq N_1$ and $N_2 \in \mathbb{N}$ such that $|b_n - b| < \epsilon \forall n \geq N_2$ Then:

$$|a_nb_n - ab| = |a_nb_n - a_nb + a_nb - ab|=|a_nb_n - ab| \leq |a_nb_n - ab| \leq |\epsilon - \epsilon| = 0$$

Are theses conclusions correct?

• For a start, you can't just choose $\varepsilon$ to be equal to some value. You need to let $\varepsilon > 0$ be arbitrary. – lokodiz Aug 29 '13 at 14:46
• @SimonC changed it... – user2051347 Aug 29 '13 at 14:52
• The inequality $|a_nb_n - ab| \leq |\epsilon - \epsilon|$ is false. – Umberto P. Aug 29 '13 at 14:53
• Notice that the third term in your equality is the same as you first term. As is the fourth term. – lokodiz Aug 29 '13 at 14:54

• $|a_n b_n - ab| \le |a_n b_n - a_n b| + |a_n b - ab|$ (why?)
• Since $a_n \rightarrow a$, $|a_n|$ can be bounded by some $M > 0$
The calculations look a bit redundant and in the last step you can't have $|\epsilon - \epsilon|$ for $|a_n b_n - ab|$. Also as a comment pointed out, you need to consider arbitrary $\epsilon > 0$ and show $|a_n b_n - ab| < \epsilon$ for sufficiently large $n$. A hint on how to proceed: If $a_n = a + \delta_n$ and $b_n = b + \gamma_n$, then $a_n b_n = ab + \delta_n b + \gamma_n a + \delta_n \gamma_n$. For given $\epsilon > 0$, you just need that $|\gamma_n a| < \epsilon / 3$, and $|\delta_n b| < \epsilon / 3$, and $|\delta_n \gamma_n| < \epsilon / 3$.